Question

4) Write the quadratic equation of the following in vertex form: vertex (-1,3) and passes through (1,-5)

Answer 1

You will have to set up a system of equations

Answer 2

Answer:

Please check the explanation.

Step-by-step explanation:

4)

If an equation representing a parabola is in vertex form such as

$y\:=a\left(x-k\right)^2+h$

then its vertex will be at (k, h).

Therefore the equation for a parabola with a vertex at (-1, 3), will have the general form

$y\:=a\left(x+1\right)^2+3$

If this parabola also passes through the point (1, -5) then we can determine the  'a ' parameter.

$-5\:=a\left(1+1\right)^2+3$

simplifying the equation

$2^2a+3=-5$

$4a+3=-5$

subtract 3 from both sides

$4a+3-3=-5-3$

$4a=-8$

Divide both sides by 4

$\frac{4a}{4}=\frac{-8}{4}$

$a=-2$

So our equation in vertex form is:

$y\:=-2\left(x+1\right)^2+3$

5)

Given the expression

$15n^2-6n$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c$

$=15nn-6n$

$\mathrm{Rewrite\:}6\mathrm{\:as\:}3\cdot \:2$

$\mathrm{Rewrite\:}15\mathrm{\:as\:}3\cdot \:5$

$=3\cdot \:5nn-3\cdot \:2n$

Factor out the common term 3n

$=3n\left(5n-2\right)$

6)

Given the expression

$2x^2+5x-10x-25$

Factor 2x²+5x: x(2x+5)

Factor -10x-25: -5(2x+5)

so the expression becomes

$=x\left(2x+5\right)-5\left(2x+5\right)$

$\mathrm{Factor\:out\:common\:term\:}\left(5+2x\right)$